Characterizing Vector Spaces without (provable) Basis in ZF Without the axiom of choice certain vector spaces cannot be proven to have a (hamel) basis in ZF alone and I am wondering whether there exists some criteria characterizing such spaces.
Here is what I know so far: Since "every vector space has a basis" implies AC (http://www.math.lsa.umich.edu/~ablass/bases-AC.pdf) every model of $ZF+\neg AC$ contains vector spaces without a basis. There are also concrete examples of vector spaces which do not have a basis in certain models of ZF such as $\mathbb{R}$ over $\mathbb{Q}$ (https://people.clas.ufl.edu/zapletal/files/lru3.pdf). However I have not yet been able to find a source giving some sufficient condition $(*)$ and a theorem of the form:

For every field $K$ and $K$-vector-space $V$ sattisfying $(*)$ it is consistent with ZF to assume that $V$ has no basis over $K$.

I strongly suspect that one can construct (via symmetric forcing) a model of ZF where

For every field $K$ and $K$-vector-space $V$ such that $K^\mathbb{N}$ embeds into $V$, $V$ has no basis over $K$.

(here $K^\mathbb{N}$ denotes the space of all functions from $\mathbb{N}$ into $K$ with pointwise addition and multiplication)
$\bullet$ Does anyone know a source confirming this statement? (If true then I imagine someone must have already proven either this or some stronger result.)
$\bullet$ Is there maybe even a more general characterization of those spaces that can't be proven to have a basis in ZF and those that can? (e.g. the criterion I've given above does not seem to cover $\mathbb{R}$ over $\mathbb{Q}$, so there must be a more general one)
All I could find on the subject was

Howard, Paul; Tachtsis, Eleftherios, On vector spaces over specific fields without choice, Math. Log. Q. 59, No. 3, 128-146 (2013). ZBL1278.03082.

but it doesn't seem to answer my question (or maybe I missed something).
 A: To my knowledge there is no such condition. Certainly not at the generality you're asking for.
The closest we can get is "infinite dimensional Banach space" (well, to be honest we can go quite a lot more, but I don't remember the exact limitation there). The reason being that under certain assumptions (most notably $\sf ZF+DC+BP/LM$, where $\sf BP$ states that every set of reals has the Baire property and $\sf LM$ states that every set of reals is Lebesgue measurable) every linear operator from a complete Polish group into a normed group is continuous, and from that we can deduce the result for any Banach space. And of course, if there is an infinite dimensional vector space, we can arrange for a discontinuous functional.
You might want to think that if $K^\Bbb N$, which might not have a basis, embeds into a vector space then that space won't have a basis either. Unfortunately it is possible for a vector space to have a basis, while some of its subspaces don't have a basis. An example can be found in If $V$ is a vector space with a basis. $W\subseteq V$ has to have a basis too?.
In general, these sort of questions are very hard since we don't have precise enough tools to deal with arbitrary vector spaces over arbitrary fields. Hopefully with recent progress in the field of symmetric extensions we can try and improve the situation there, but frankly there is a long long way to go.
